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Tom [10]
3 years ago
5

Ariel completed the work below to show that a triangle with side lengths of 9, 15, and 12 does not form a right triangle. mc018-

1.jpg Is Ariel’s answer correct? No, Ariel should have added 122 and 1
Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

Ariel answer is wrong.

Step-by-step explanation:

Given that Ariel had done a Math problem for a triangle with sides

9,15 and 12.  She concluded that these sides are not that of a right triangle.

To check whether Ariel is right:

Greater side = 15

Hence greatest angle will be opposite 15

Let us check pythagorean theorem

smaller sides are 9 and 12

Square and add them

9 square +12 square = 81+144 = 225 =15 square

We find that Pythagorus theorem is satisfied

Hence it is concluded that the triangle is right angled and Ariel was wrong

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If its 10/16 what is answer as a decimal
Vedmedyk [2.9K]

Answer:

0.625 is a decimal and 62.5/100 or 62.5% is the percentage for 10/16.

Step-by-step explanation:

Please mark brainliest and have a great day!

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3 years ago
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Try to get to every number from 1 to 10 using four 4's and any number of arithmetic operations (+, −, ×, ÷). You may also you pa
stepan [7]

Answer:

Step-by-step explanation:

1. 4/4+4-4=1

2. 4/4+4/4=2

3. 4+4/4-4=3

4. 4 × (4 − 4) + 4=4

5. (4 × 4 + 4) / 4=5

6. 44 / 4 − 4=6

7. 4+4-4/4=7

8. 4+4+4-4=8

9. 4+4+4/9=9

10. 44 / 4.4=10

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3 years ago
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How do you write the decimal expansion for 6/11
xxMikexx [17]

Divide 6 by 11

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Answer is A


5 0
3 years ago
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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



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kherson [118]
After 7 days she would have eventually found 63 cents.
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